Exercise 2.8.30* (Period of the decimal expansion of a rational number)

Proof. Here $q>1$. Let $k$ be the order of $10$ modulo $q$. The order is well defined, because $10\wedge q=1$.

Consider the rational $a∕q$. The Euclidean division gives integers $m,r$ such that $a=\mathit{mq}+r$, where $0\le r<q$. Moreover $r\ne 0$, otherwise $q\mid r$, and $q\wedge r=q>1$. So $0<r<q$. Then

The decimal expansion of $a∕q$ if composed of the decimal expansion of $m=\lfloor a∕q\rfloor$, followed by a dot and the decimal expansion of $r∕q<1$. Now we prove that the decimal expansion of $r∕q$ is periodic with least period $k$.

Define inductively the sequence ${(r_i)}_{i\in \mathbb{N}}$ by $r_0=r$, and for all $i\in \mathbb{N}$,

where $a_{i+1},r_{i+1}$ are the quotient and remainder of the division of $10r_i$ by $q$. Note that $a_{i+1}q<10r_i<10q$, thus $0\le a_{i+1}<10$, so the $a_i$ are decimal digits.

(The $a_i$ are the digits of the result of the usual algorithm of division seen in primary school. For instance, for $7∕13$,

Using (1), we obtain by induction that

Then

thus

This show that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{a}_{i}10^{-i}=0.a_1{a}_2\cdots a_n$ is a default approximation to within $10^{-n}$ of $r∕q$: the algorithm of primary school is not so bad...

From (1) we deduce $r_{i+1}\equiv 10r_i(\mathit{mod}q)$ for all $i\in \mathbb{N}$, therefore, by immediate induction, for all $n\in \mathbb{N}$,

Since $10^k\equiv 1(\mathit{mod}q)$, $r_k\equiv r_0=r\mathit{mod}q$, where $0\le r_0<q,0\le r_k<q$, hence $r_k=r_0$. The unicity of the remainder shows that $r_{k+1}=r_1,r_{k+2}=r_2$, and more generally $r_{k+j}=r_j$ for all $j\in \mathbb{N}$. This shows that $k$ is a period of the sequence ${(r_i)}_{i\in \mathbb{N}}$. Moreover $r_{k+j}=r_j$ implies by unicity of the quotient that $a_{k+j+1}=a_{j+1}$, so that $k$ is a period of the decimal expansion of $r∕q$ (or $a∕q$). (Note that the periodicity begins with the first digit, and not after some time.)

We must show now that $k$ is the least period.

Let $h$ be any period of the decimal expansion of $r∕q$. Then

where $a=a_{1}10^{h-1}+\cdots +a_h$ is an integer. Then

Then $r(10^h-1)=\mathit{aq}$, so $q\mid r(10^h-1)$, where $r\wedge q=1$. Therefore $q\mid 10^h-1$, so $10^h\equiv 1(\mathit{mod}q)$. The order of $10$ is $k$, thus $k\mid h$. This proves that $k$ is the least period of the decimal expansion of $r∕q$ or $a∕q$. □